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Mariana [72]
3 years ago
7

Assume that you have 1 mL of a solution of amylase (an enzyme) at a concentration of 15 mg protein/mL. Calculate the volume of d

iluting buffer that you would have to add to 1.0 mL of the amylase stock solution if you wished the final concentration of the solution to be 345 µg protein/mL.
Biology
1 answer:
lidiya [134]3 years ago
8 0

Answer:

42,5 mL

Explanation:

We need to use the serial dilution formula beacuse we start with a stock concentrate solution and we need to prepare a new less concentrated one.

DF=\frac{Vi}{Vf}

<u>DF in the dilution factor, Vi is the initial volume and Vf is the final volume.</u>

The first step is to have the same measurment unit so we need to convert 345 µg to mg.

we know that 1 µg equals 0,001 g, hence:

345 µg = 0,345 mg

now the final volume is 0,345 mg  protein/ mL and the inital volume is 15mg protein/mL, both of them are in the same unit so we can use the formula

DF= \frac{15mg protein/mL}{0.345mg protein/mL}

DF= 43,5 mg protein/ mL

Now since the question said that we already have 1.0mL of the amylase stock solution we need to subtract that 1.0mL to the 43,5 mg protein/mL

43,5mL-1,0mL = 42,5 mL

So, we need 42,5 mL of diluting buffer if we want a final concentration of 345 µg protein/mL (0.345 mg protein/mL)

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